1. Two Sum

1. Two Sum

Description:

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

思路

假设这些数排好序了， 那么我们只需要使用双指针即可在O(n)的时间内解决此问题。代码如下:

class Solution {
public:
    static bool cmp(const pair<int, int>& a, const pair<int, int>& b) {
        return a.first < b.first;
    }
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<pair<int, int> > num;
        for(int i=0; i<nums.size(); i++) {
            num.push_back(make_pair(nums[i], i));
        }
        sort(num.begin(), num.end(), cmp);
        int i = 0, j = nums.size()-1;
        vector<int> res;
        while(i < j) {
            int a = num[i].first;
            int b = num[j].first;
            if(a+b > target) j--;
            else if(a+b < target) i++;
            else {
                res.push_back(num[i].second);
                res.push_back(num[j].second);
                break;
            }
        }
        return res;
    }
};